TC3 → Stan Brown → Statistics → Chapter 9 Practice Quiz
revised 10 Apr 2013 (What’s New?)

Practice Quiz with Solutions: Chapter 9 (25 min)

Before looking at these solutions, please work the practice quiz.

Disclaimer: This quiz is representative of the level of difficulty you should expect, but it doesn’t include every single topic from the week’s work. The real quiz may include some other topics and may skip some that are in this practice quiz. (The real quiz also may not word questions in the same way as the practice quiz. You should focus on the concepts, not a particular form of words.)

See also: How to Take a Math Test

These solutions show about the same level of work I expect from you, though I often add some extra commentary. Please see Show Your Work for the what, why, and how.

1(points: 5)  Lito-Fray was monitoring the percent salt content of batches of potato chips, and found the following results in a random sample of 11 batches:

1.6   1.4   1.6   0.9   1.3   1.4   1.5   1.4   1.2   1.3   1.7

Estimate the average salt content of all chips, by constructing a 95% confidence interval in endpoint form. Interpret your answer in English.

Solution: This is numeric data, one population mean with population standard deviation unknown: Case 1 on Inferential Statistics: Basic Cases.

Make sure the requirements are satisfied. Since the sample size is under 30, you must check that the sample data are normally distributed with no outliers. Your two test screens should look like this:

MATH200A/2 box-whisker plot showing absence of outliers    MATH200A/5 normal probability plot showing r=.9624

You don’t need to copy the screens, but you should say that you used MATH200A part 2 and found no outliers, and you used MATH200A part 4 and found a normal distribution (r=.9624, crit=.923, r>crit).

Since you don’t know the standard deviation of the population, use a T-Interval to compute the confidence interval.

TInterval input screen; see text    TInterval output screen; see text

And you show your work as

TInterval inputs: List:L1, Freq:1, C-Level:.95

outputs: (1.2423,1.5395), x̅=1.39; s=0.22; n=11

Answer: We have 95% confidence that the average salt content of all chips is between 1.24 and 1.54 percent.

Remark: By convention, we round means and standard deviations to one more decimal place than the original data. It’s not the end of the world if you do it differently, but following the convention is usually a good choice.

Common mistake: Make sure that your confidence interval refers to the population, not to the sample. Any reference to “11 batches” in your confidence interval is wrong. You’re not estimating anything about the 11 batches; you know their mean exactly.

Alternative solution: If you prefer, you can do 1-Var Stats before TInterval. If you do that, the calculator pastes the values into the TInterval screen for you:

1-VarStats output screen    TInterval input screen; see text

The output screen would be the same, of course. You show your work as

1-VarStats L1

TInterval x̅:1.3909, s:0.2212, n:11, C-Level:.95

outputs: (1.2423,1.5395)

2(points: 1)  In World War II, a prisoner of war flipped a coin 10,000 times and recorded 5,067 heads. What number is the point estimate for proportion of heads?

Answer: p̂ = 5067/10000 = 0.5067

Remark: Students seem to have an awful lot of trouble with the term “point estimate”. Don’t make it harder than it needs to be: the point estimate for the population mean (or proportion, standard deviation, etc.) is very simply the sample mean (or proportion, standard deviation, etc.).

3(points: 2½)  Last week, 42% of TV viewers watched American Idol. You wonder what percent will watch it this week. You want to be 90% confident of your answer, with a margin of error no greater than ±3.5%. How large must your random sample be?

Solution: This is Case 2, one population proportion, a sample size problem. You have a prior estimate of 42% or 0.42 for the true proportion. Use MATH200A part 5. Enter

Data Type: 3:Binomial,  p̂: .42,  E: .035,  C-Level: .9

and read off the answer: n ≥ 539

Common mistake: Convert percentages to decimals carefully! 3.5% is 0.035, not 0.35 or 3.5.

Common mistake: Always use a prior estimate when you have one. If you use the conservative figure of p̂ = 0.5, you get a sample size of 553, which is larger than necessary.

Alternative solution: n=p times 1 minus p times square of z of half alpha, over E squared If you don’t have the program, or if you prefer to work out the solution, use the formula at right. You have a prior estimate of 42% or 0.42 for the true proportion, so you use p̂ = 0.42 and 1−p̂ = 0.58. E is 3.5% or 0.035 (given).

For z(α/2), start with 1−α = 0.90; therefore α = 0.10 and α/2 = 0.05. Now you need z0.05, which is invNorm(1−0.05) or about 1.6449. (From Chapter 7, z0.05 is the z score that divides the normal curve so that the area to left is 0.05.) Substituting,

n = .42 * (1−.42) * (invNorm(1−0.05)/0.035)² = 538.0166

Since the sample size must always be a whole number, greater than or equal to the computed n, your sample must be at least 539 people.

See also: How Big a Sample Do I Need?

Common mistake: When you get the answer of 538-point-something, don’t stop there. Sample sizes are always rounded up to the next whole number.

4(points: 2½) You’re planning a study of vacation spending. From a past study, you have reason to believe that spending plans are normally distributed with σ = $600. How many people will you need to ask, if you want a 98% confidence level and a margin of error no more than $250?

Solution: This is Case 0, numeric data with known s.d. of the population. Use MATH200A part 5 again. Enter

Data Type: 1:Num known σ,  σ: 600,  E: 250,  C-level: .98

and the answer is n ≥ 32

Alternative solution: n equals z of alpha over 2, times sigma, over E, all squared If you don’t have the program, use the formula at right. Here σ = 600, E = 250, and you have to find zα/2. (From Chapter 7:  zα/2 is the z score that divides the normal curve so that the area to left is α/2.)

C-Level = 1−α = .98 ⇒ α = .02 ⇒ α/2 = .01

zα/2 = z.01 = invNorm(1−.01) ≅ 2.3263

n = ( invNorm(1−.01) * 600 / 250 )² = 31.1725 → 32 people

Common mistake: The answer is not 31 people because you don’t round sample size. Rather, you interpret it as “I need at least 31.1725 responses”, which means 31 won’t be enough but 32 is adequate.

5(points: 4)  You’ve read that two thirds of Americans believe in angels. You want to see whether that’s true at TC3, so you take a random sample of 100 TC3 students and you find that 61 of them believe in angels. Construct and interpret a 95% confidence interval about the proportion of all TC3 students who believe in angels.

Solution: This is binomial data for one population: Case 2.

1-PropZInt: x=61, n=100, C-Level=.95

Outputs: (.5144,.7056); p̂=.61

Check the requirements, using p̂ from the calculator.

Is np̂(1−p̂)≥10?  p̂ = 0.61 and n = 100. np̂(1−p̂) = 100*.61*(1−.61) = 23.79; check.

Is 20n≤N? 20*100 = 2000, and the TC3 student population numbers greater than 2000; check.

Answer: We’re 95% confident that 51.4% to 70.6% of all TC3 students believe in angels.

Common mistake: Again, your statement needs to be about the population (all TC3 students), not about the sample (100 students). There’s no estimation related to the 100 students, because you know that exactly 61 of them believe in angels.

Remark: It’s possible that two thirds of TC3 students believe in angels. We can't really tell because 66.7% is in the 95% confidence interval.

What’s New

This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit www.tc3.edu/instruct/sbrown/ to report errors or ask to copy it.

For updates and new info, go to http://www.tc3.edu/instruct/sbrown/stat/